# Plot $g(x)=e^{\pi'(x)}$

Assuming the Riemann Hypothesis,

$$\pi(x)=R(x)-\sum_\rho R(x^\rho),$$

where $$\pi(x)$$ is the prime counting function and $$R(x)$$ is the Riemann prime counting function.

What does a plot of $$g(x)=e^{\pi'(x)}$$ look like?

Another option might be using:

$$\quad\Pi'(x)=\sum\limits_{n=2}^\infty\frac{\Lambda(n)}{\log(n)}\,\delta(x-n)\quad\text{(derivative of Riemann's prime-power counting function)}.$$

Define $$f(x)=e^{\Pi'(x)}.$$ What does the plot of $$f(x)$$ look like?

Edit: Assuming the Riemann Hypothesis $$\pi(x)=Li(x)+O(\sqrt{x}\ln(x)).$$

I can take the derivative of the first term $$Li(x)$$ but I'm not sure how to take the derivative of the second term.

• No need to assume the RH. Replace $\pi(x)$ by $\psi(x)$ and its corresponding explicit formula $\psi(x) = (x-\sum_\rho x^\rho/\rho-C-\sum_{k=1}^\infty x^{-2k}/(-2k))1_{x > 1}$ which is much easier, $\psi'(x) = \sum_n \Lambda(n) \delta(x-n)$ is a distribution, $e^{\delta(x)}$ and $e^{\psi'(x)}$ don't make any sense. If your question is if $(1-\sum_\rho x^{\rho-1}-\sum_{k=1}^\infty x^{-2k-1})1_{x > 1}$ (dont forget the $1_{x > 1}$) converges to the distribution $\psi'(x)$ then yes this is the Weil explicit formula. It doesn't converge in the function sense. – reuns May 5 at 16:50
• @Ultradark I believe the answer I posted to a related question at math.stackexchange.com/q/3186899 perhaps provides some insight. – Steven Clark May 5 at 19:58